= Solution
\b[Yes.] Let $b$ be the vertex of $(\Delta^3)'$ corresponding to the whole tetrahedron. Its link in $M$ is the 1-skeleton of the barycentric subdivision of $\partial\Delta^3$. This connected graph has one vertex for each of the 14 nonempty proper faces and 36 edges, so
$$
H_2(|M|,|M|\setminus\{b\})
\cong\widetilde H_1(\operatorname{link}(b))
\cong\mathbb Z^{36-14+1}
=\mathbb Z^{23}.
$$
No other point has this local homology rank. At the other vertices, corresponding respectively to faces of size $1,2,3$, the first Betti numbers of the links are $6,4,6$. A point in the interior of an edge is incident with at most six triangles, so its local $H_2$ has rank at most five; a point in a triangle interior has local $H_2\cong\mathbb Z$.
A homeomorphism preserves <local homology>, so it must send the unique point with local $H_2$ of rank $23$ to itself. Therefore every self-homeomorphism fixes $b$, which is the <fixed barycentre of the barycentric tetrahedral two-skeleton>.
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